Crankshaft is subjected to pulls F1 and F3 parallel to z axis and F2 and F4 parallel to y axis .What are bearring reactions at A and B if the pulls...

The total reaction in A is parallel to the total reaction
in B, it is equal in magnitude with B and it is opposite in
direction.

The 2 total reactions form a couple, since
F1-F3, F2-F4 form couples when the forces are of
equal magnitude.

We'll write the equations of
equilibrium:

Sum of torques of z axis =
0

+F2(a+b)-F4(a+3b)+By(2a+3b) =
0

-Ay(2a+3b) - F2(a+2b) + F4*a =
0

Sum of torques of y axis =
0

-F1*a + F3(a+2b) - Bz(2a+3b) =
0

Az(2a+3b) + F1(a+3b) - F3(a+b) =
0

We'll consider F1=F2=F3=F4=F and we'll open the brackets
and we'll have:

Fa + Fb - Fa - 3Fb + 2aBy + 3bBy =
0

We'll eliminate and combine like
terms:

-2Fb + 2aBy + 3bBy = 0
(1)

- 2aAy - 3bAy - Fa - 2bF + Fa =
0

We'll eliminate and combine like
terms:

- 2aAy - 3bAy - 2bF = 0
(2)

-Fa + Fa + 2bF - 2aBz -3bBz =
0

2bF - 2aBz -3bBz = 0
(3)

2aAz+3bAz + Fa+ 3bF - Fa + Fb =
0

We'll eliminate and combine like
terms:

2aAz+3bAz + 4bF = 0
(4)

Bz =
2bF/(2a+3b)

By =
2bF/(2a+3b)

Az =
-2bF/(2a+3b)

Ay =
-2bF/(2a+3b)